RI Carboyxlic acids and Derivatives Tutorial - Answers to Discussion Question

4 Suggested Answers to Carboxylic Acid & Derivatives Tutorial 1(a) Decreasing acid strength: ethanoic acid > phenol > ethanol In general, HA + H2O ⇌ H3O+ + A– The acidity of a compound depends on the relative stability of its conjugate base anion (A–). The more stable the conjugate base, the more acidic the compound will be. In the ethanoate ion (CH3CO2–), the negative charge on oxygen is dispersed over the two highly electronegative oxygen atoms resulting in two equivalent resonance structures. Hence, CH3CO2 ion is resonance-stabilised to a larger extent than C6H5O– ion. In the phenoxide ion (C6H5O–), the p-orbital containing the lone pair of electrons on the O atom overlaps with the -electron cloud of the benzene ring so that the negative charge on O delocalises into the benzene ring. The dispersal of negative charge stabilises C6H5O– ion but this resonance stabilisation is not as great as that in the CH3CO2 ion. In the ethoxide ion (CH3CH2O–), the electron-donating alkyl (ethyl) group intensifies the negative charge on O atom, which destabilises CH3CH2O– ion. Therefore, CH3CH2O– is the least stable. (b) Decreasing acid strength: trifluoroethanoic acid > trichloroethanoic acid > chloroethanoic acid > ethanoic acid In general, HA + H2O H 3O+ + A– The acidity of a compound depends on the relative stability of its conjugate base anion (A–). The more stable the conjugate base, the more acidic the compound will be. Compared to CH3COO ion, the CF3COO, CCl3COO and ClCH2COO ions are more stable due to the presence of electron-withdrawing F or Cl group(s) to disperse the negative charge and hence stabilising the anion. CCl3COO ion is more stable than ClCH2COO ion as it has 2 more electron-withdrawing Cl groups to disperse the negative charge. Since F is more electronegative than Cl, the F group is more electron-withdrawing than the Cl group and the negative charge is dispersed to a greater extent in CF3COO ion than in CCl3COO ion. Hence CF3COO ion is more stable than CCl3COO ion. 2 Decreasing ease of hydrolysis: C6H5COCl > C6H5CH2Cl > C6H5Cl The carbon of the acyl group in C6H5COCl has a higher + charge (or is more electron deficient) as it is bonded to two electronegative atoms (O and Cl). The carbon bonded to the chlorine atom in C6H5CH2Cl has lower + charge (or is less electron deficient) as it is bonded to only one electronegative atom (Cl). Hence, C6H5COCl can attract nucleophiles more easily and is more susceptible to nucleophilic attack as compared to C6H5CH2Cl. In addition, the carbon of the acyl group in C6H5COCl, being sp2 hybridised and trigonal planar, provides less steric hindrance during nucleophilic attack compared to the carbon bonded to the chlorine atom in C6H5CH2Cl, which is sp3 hybridised and tetrahedral. Thus, C6H5COCl undergoes hydrolysis with ease with water (a weak nucleophile) while C6H5CH2Cl requires a stronger nucleophile OH− under heating.

5 C6H5Cl is the least susceptible to hydrolysis. This is because the p orbital containing the lone pair of electrons on the Cl atom overlaps with the π electron cloud of the benzene ring, resulting in a lone pair of electrons in the p orbital of Cl delocalising into the benzene ring. As a result, the C-Cl bond has partial double bond character. Since the bond is strengthened, the cleavage of this bond (which is necessary during hydrolysis) is made very difficult. 3 (a) Test: Add aq Na2CO3 (or aq NaHCO3) to each compound in a test-tube. Observation: For C 6H5COOH: Effervescence observed, CO2 gas evolved formed white ppt with limewater/aq Ca(OH)2. For C 6H5OH: No effervescence / no gas is evolved. Other tests: Neutral FeCl3(aq): Violet complex observed for phenol. No violet complex for benzoic acid. Or aqueous Br2: Phenol decolourises orange Br2 and a white ppt is formed. No decolourisation of Br2 and no white ppt for benzoic acid. (b) Test: Add aq AgNO3 to each compound in a test-tube. Observation: For CH 3CH2COCl: White ppt (AgC l) formed and white fumes (HCl) evolved. For C lCH2CH2COOH: No white ppt and no white fumes. Note: Aq. Na2CO3 should not be used as a distinguishing test between acid chlorides and carboxylic acids. Acid chlorides such as CH3CH2COCl undergo hydrolysis readily in aqueous medium, under room temperature conditions. The products, HCl and carboxylic acid, will react with aq. Na2CO3 to produce CO2. (c) Test: Add NaOH(aq) to each compound in a test-tube and heat in a hot water bath. Then add aq I2. OR Add NaOH(aq) and I2(aq) to each compound in a test-tube and heat in a hot water bath. Observation: For CH 3COOCH(CH3)2: Yellow ppt of CHI3 For CH 3COOCH2CH2CH3: No yellow ppt formed.

6 (d) Test: Add 2 drops of aq KMnO4 and aq H2SO4 to each compound in a test-tube and heat in a hot water bath. Observation: For HCOOH: Purple KMnO4 decolourises. Effervescence observed, CO2 gas evolved formed white ppt with aq limewater/Ca(OH)2. For CH3COOH: Purple KMnO4 is not decolourised. Note: HCOOH also gives a positive test with Tollens’ reagent & Fehling’s solution. 4 (a) CH3OH + ClOHO (b) CH3OHOHOONa+ (c) CO2 + OHOOHO + ClOHO 5 (a) (b) Note: HCl with ZnCl2 can also be used for the first step. 6 (a) Evidence Deduction F, C8H8O2 F has a high C:H ratio of 1:1 and > 6 C atoms  F likely contains benzene ring F insoluble in water presence of large non-polar group such as benzene ring in F F dissolves in NaOH(aq) F undergoes acid-base reaction  F contains acidic functional group  F is either a phenol or a carboxylic acid F reacts with 2,4-DNPH F undergoes condensation with 2,4-DNPH  F is a carbonyl compound HCl(aq), heat/heat under reflux

7 F does not react with Fehling’s solution F is not oxidised by Fehling’s solution  F is not an aliphatic aldehyde  F is either an aromatic aldehyde or a ketone F + Br2(aq)  G (C8H5O2Br3) F undergoes electrophilic substitution with Br2(aq)  F contains a strongly activated benzene ring  F is a phenol G has 3 H atoms substituted with 3 Br atoms  G has Br substituted at the 2,4 and 6 positions with respect to the phenol OH group. F + I2/ OH−(aq) then H+  H F undergoes oxidation by alkaline I2(aq) (positive iodoform test) followed by acid-base reaction to form H  F is a methyl ketone  H is a carboxylic acid with 1 C less than F H dissolves in both NaOH(aq) and Na2CO3(aq). H undergoes acid-base reaction with NaOH and Na2CO3  H is a carboxylic acid (confirmed) (b) (i) (ii) Note: An extra mol of OH− is required to neutralise the acidic phenol group. (iii) F G H

8 7 Ans: B (2) In D2O solvent, the phenol and COOH groups can lose H+ to form the respective conjugate bases. Using the COOH group as an example: The carboxylate anion can then undergo basic hydrolysis with D2O to form RCOOD as the deuterated solvent is present in large excess: 8(a) Br O HO OH OHO OBr C OH Nstep 1 step 2 step 3 C CN OH N Step 1: HCN, trace of KCN Step 2: ethanolic KCN, heat/heat under reflux Step 3: dilute HCl, heat/heat under reflux [Note : Step 1 and 2 involving cyanide could be carried out in any order] (b) (i) cis-trans isomerism C C COOH H COOH H C C COOH COOH H H (ii) excess concentrated H2SO4, heat (or Al2O3, heat) (iii) The pKa values suggest that K, having a low first pKa coupled with a high second pKa produces the more stable mono-anion. The first p Ka is much lower implying that K loses its first proton more readily to form the mono-anion and it is therefore also more difficult for the mono-anion of K to accept H+. The second pKa is much higher, implying it is more difficult for the mono-anion to lo